Classification of elements and periodicity in properties practice · Mixed difficulty

Option D (1509) is correct because it correctly calculates the second ionization energy based on the given ratio and the sum of the first and second ionization energies.

Option C is correct because atomic size decreases across a period due to increasing nuclear charge, but Neon is larger than Fluorine due to increased electron-electron repulsion in the fully filled electron configuration.

Option D (Fluorine) is correct because Fluorine is the most electronegative element.

Option B (Chlorine) is correct because chlorine has the highest electron affinity among the given options due to its smaller size and higher effective nuclear charge compared to sulphur, and fluorine's electron affinity is lower than chlorine's due to interelectronic repulsions in its small 2p orbitals.

Option B (Rb) is correct because Rubidium (Rb) is the first element in the fifth period of the periodic table.

Option B (d-block) is correct because d-block elements have variable oxidation states and form complexes, making them good catalysts.

Option C (${Cs}$) is correct because Cesium is the lightest liquid metal at or near room temperature.

Option D ($B < S < P < F$) is correct because ionization energy generally increases across a period, with exceptions due to half-filled orbitals (P).

Option A (Iodine) is correct because Iodine, despite having a lower atomic weight than Tellurium, was placed after Tellurium in Mendeleev's periodic table due to its chemical properties.

Option A (actinoids and group 4) is correct because element 101 is an actinoid (Mendelevium) and element 104 is in group 4 (Rutherfordium).

Option A (Fifth) is correct. The electronic configuration ends in $3d^3 4s^2$, indicating that it's a d-block element in the 5th group (3+2 = 5).

Option C (Density : Li < Na < K < Rb) is correct because the density of K is actually less than Na due to its larger size and weaker metallic bonding.

Option B (+526 kcal) is correct because it represents the sum of the first and second ionization energies, which is the total energy required to remove two electrons from Mg.

Option B (IB) is correct because Group IB (or Group 11) of the Periodic Table contains coinage metals like copper, silver, and gold.

Option D (Halogen family) is correct because element 117 would be in Group 17, the Halogen family.

Option B (electronic configuration) is correct because the modern periodic law states that properties are periodic functions of electronic configuration.

Option A (Increasing atomic number) is correct because elements are arranged in order of increasing atomic number in the long form of the periodic table.

Option C ($F_2$) is correct because both $O_2^{2-}$ and $F_2$ have 18 electrons.

Option B (Halogens) is correct because elements with atomic numbers 9 (Fluorine), 17 (Chlorine), 35 (Bromine), 53 (Iodine), and 85 (Astatine) are all halogens.

The correct answer is A (Have diagonal relationship) because Lithium (Li) and Magnesium (Mg) exhibit similar chemical properties due to their diagonal relationship in the periodic table.

Option B (O) is correct because oxygen has the highest electronegativity among the given elements.

Option A is correct because the acidic strength of oxides increases with the electronegativity and oxidation state of the central atom: Cl2O7 > SO2 > P4O10.

Option C (O, S) is correct because sulfur is below oxygen in the same group, so it has a larger atomic radius.

Option C is correct because Helium is placed in the 's' block, not the 'p' block, and its differentiating electron enters the 's' orbital.

Option B is correct because both the assertion and reason are true, but the reason (stable half-filled configuration) is not the correct explanation for the assertion (electronegativity). Electronegativity is primarily related to nuclear charge and atomic size.

Option A ($\mathrm{C}{\mathrm{l}_2}{\mathrm{O}_7} > S{\mathrm{O}_2} > {\mathrm{P}_4}{\mathrm{O}_{10}}$) is correct because the acidic strength of oxides increases with the electronegativity and oxidation state of the central atom.

Option A (Inner electrons) is correct because inner electrons shield the outer electrons from the full nuclear charge.

Option B (Sn) has the lowest ionization energy because ionization energy generally increases up a group, and Sn is below Pb, Si, and C in Group 14.

Option D (None) is correct because the second ionization energy is always greater than the first ionization energy for any element or ion.

Option A ($\mathrm{A}_2 \mathrm{O}_3$) is correct because among the given elements, Boron has the smallest atomic radius and forms $\mathrm{B}_2 \mathrm{O}_3$. Aluminum also forms $\mathrm{Al}_2 \mathrm{O}_3$.